杭电2056Rectangles(未解决)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2056

该题Output Limit Exceeded，可能是考虑情况太少了！

下面的代码Output Limit Exceeded了

#include <cstdlib>
#include <iostream>
#include <stdio.h>
using namespace std;

int main(int argc, char *argv[])
{
    double a[8]={0};
    //int n=2;
    while(1)
    {
      double x=0,y=0;
      double area=0;
      for(int i=0;i<8;i++)
         cin>>a[i];
      if(a[0]>a[4])
      {
        x=a[6]-a[0];
        y=a[7]-a[1];
        area=x*y;
        printf("%.2lf\n",area);
      }
      if(a[0]<a[4])
      {
        x=a[2]-a[4];
        y=a[3]-a[5];
        area=x*y;
        printf("%.2lf\n",area);
      }
        
    }
    system("PAUSE");
    return EXIT_SUCCESS;
}

 解决它~~~~~~~~~

下面的代码Time Limit Exceeded了

#include <cstdlib>
#include <iostream>
#include <stdio.h>
using namespace std;

int main(int argc, char *argv[])
{
    double a[8]={0};
    //int n=2;
    while(1)
    {
      double x1=0,y1=0,x2=0,y2=0;
      double area=0;
      for(int i=0;i<8;i++)
         cin>>a[i];
      if(a[0]>a[4])
        x1=a[0];
      else
       x1=a[4];
      if(a[1]>a[5])
        y1=a[1];
      else
        y1=a[5];
      if(a[2]>a[6])
        x2=a[6];
      else
        x2=a[2];
      if(a[3]>a[7])
        y2=a[7];
      else
        y2=a[3];
      if(x2>x1&&y2>y1)
      {
        area=(x2-x1)*(y2-y1);
        printf("%.2lf\n",area);
      }
       
        
    }
    system("PAUSE");
    return EXIT_SUCCESS;
}

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